By [1]}, \(E\)  is simply connected and coincides with \(\exp \mathfrak {e}\) . In particular, \(E\)  is integral. Since \(\mathfrak {n}=[\mathfrak {g},\mathfrak {g}]\cap \mathfrak {r}\)  (see the proof of Proposition REF ), we have \(\mathfrak {e}\subset \mathfrak {n}\)  and hence \(E\subset N\) . Since \(N\)  is nilpotent, so is \(E\) .
